Chapter 9, Problem 69RE

Solution

To identify: The expression use summation formulas to evaluate ${\displaystyle \sum _{k=1}^{25}\left(k^2-3k+4\right)}$

The expression is $4650$

Given information:

The given expression is ${\displaystyle \sum _{k=1}^{25}\left(k^2-3k+4\right)}$ .

Explanation:

Consider the given information.

It is given that ${\displaystyle \sum _{k=1}^{25}\left(k^2-3k+4\right)}$

Write the sum using the sum properties.

  $S={\displaystyle \sum _{k=1}^{25}\left(k^2\right)-{\displaystyle \sum _{k=1}^{25}\left(3k\right)}}+{\displaystyle \sum _{k=1}^{25}4=}{\displaystyle \sum _{k=1}^{25}{k}^2-3{\displaystyle \sum _{k=1}^{25}\left(k\right)}}+{\displaystyle \sum _{k=1}^{25}4\cdot 25}$

Use the summation formula.

  $\begin{array}{l}{\displaystyle \sum _{k=1}^n\left(k^2\right)=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}\\ {\displaystyle \sum _{k=1}^n\left(k\right)=\frac{n\left(n+1\right)}{2}}\end{array}$

  $\begin{array}{c}S=\cdot \frac{25\left(25+1\right)\left(2\cdot 25+1\right)}{6}-\frac{3\cdot 25\left(25+1\right)}{2}+100\\ S=5525-975+100\\ S=4650\end{array}$